Question

Solve the equations and check your solutions. If there is no solution, say no 4x/x+1=4x-3/x-2

Answer 1

Does it look like this:
\(\dfrac{4x}{x+1}= \dfrac{4x-3}{x-2}\)

Answer 2

If so, start by cross multiplying.

Answer 3

yes

Answer 4

4x/2(x+1)-4x-3/x-2 4x/x+1-4x-3/x-2=0
4x(x0^2-4x-3(x+1)/x+1(x-2)x-2(x+1)
4x^2-8x-4x6+x-5

Answer 5

FYI: Using the equation editor will make your equations more clear...
I don't think you got the cross multiplication correct. Here is an example.
\(\dfrac{x}{y} = \dfrac{a}{b}\)
\(xb=ya\)

Answer 6

ok

Answer 7

got figure out how to use the equation editor

Answer 8

Your equation should look like this after you cross multiply:
\(4x^2-8x=4x^2-3x+4x-3\)

Answer 9

This site uses \(LaTeX\), I personally find that much easier to use. To see the code that someone else used, right click \(\rightarrow\) "Show Math As" \(\rightarrow\) "Tex Commands".

Answer 10

ok

Answer 11

\(4x^2-8x=4x^2-3x+4x-3\)
Is written as:
`\(4x^2-8x=4x^2-3x+4x-3\)`

Answer 12

Back to your question, do you see how I did the cross multiplication?

Answer 13

yes

Answer 14

Great! Let me know if you need any more help.

Answer 15

\(4x^2-8x=4x^2-3x+4x-3/)

Answer 16

You're slash on the right end is facing the wrong direction...
\(4x^2-8x=4x^2-3x+4x-3\)

Answer 17

My grammar/spelling gets bad after 11pm. Sorry.

Answer 18

ok

Answer 19

From here, combine like terms.

Answer 20

4x^2-8x=16x^2-3x+3

Answer 21

Use inverse operations: addition<->subtraction:
\(\cancel{4x^2-4x^2}-8x=\cancel{4x^2-4x^2}-3x+4x-3\)

Answer 22

ok

Answer 23

there is no solution